Convolution Theorem (Laplace)

Q: What are common mistakes with the Convolution Theorem (Laplace) formula?

Confusing convolution f*g with the pointwise product f(t)g(t). Forgetting that the theorem only applies if the transforms F(s) and G(s) exist for the same region of convergence.

Q: What is a real-world example of the Convolution Theorem (Laplace) formula?

In signal processing, the output of a system is the convolution of its input signal and its impulse response; this theorem allows us to find the output using multiplication in the s-domain.

Q: What are some study tips for the Convolution Theorem (Laplace) formula?

The convolution f * g is defined as the integral from 0 to t of f(τ)g(t-τ) dτ. Remember that convolution is commutative, meaning f * g = g * f.

Core idea

Overview

This theorem provides a powerful method for finding inverse Laplace transforms of products of functions by using the convolution integral.

When to use: Essential for solving non-homogeneous differential equations and analyzing linear time-invariant (LTI) systems.

Why it matters: It converts the complex operation of convolution in the time domain into simple algebraic multiplication in the frequency (s) domain.

Symbols

Variables

F(s)G(s) = L{f * g}, F(s) = F(s), G(s) = G(s)

F(s)G(s)

L{f * g}

Transform of the convolution

F(s)

Transform of the first function

G(s)

Transform of the second function

Walkthrough

Derivation

Derivation/Understanding of Convolution Theorem (Laplace)

This derivation demonstrates that the Laplace transform of the convolution of two functions is equivalent to the product of their individual Laplace transforms.

The functions f(t) and g(t) are piecewise continuous on [0, ∞) and of exponential order.
The Laplace transforms F(s) = ℬ{f(t)} and G(s) = ℬ{g(t)} exist.
The order of integration can be interchanged (Fubini's Theorem applies).

1

Start with the definition of the Laplace transform of a convolution:

We begin by applying the definition of the Laplace transform to the convolution of two functions, f(t) and g(t), which is itself an integral.

L {f * g} = \int_{0}^{\infty} e^{- s t} (\int_{0}^{t} f (τ) g (t - τ) d τ) d t

2

Change the order of integration:

The region of integration is 0 ≤ τ ≤ t < ∞. By changing the order of integration, we rewrite the limits to integrate with respect to t first, then τ.

L {f * g} = \int_{0}^{\infty} \int_{τ}^{\infty} e^{- s t} f (τ) g (t - τ) d t d τ

3

Perform a substitution in the inner integral:

Let u = t - τ, so t = u + τ and dt = du. This substitution transforms the inner integral into a form that resembles a Laplace transform.

L {f * g} = \int_{0}^{\infty} f (τ) e^{- s τ} (\int_{0}^{\infty} e^{- s u} g (u) d u) d τ

4

Recognize the Laplace transforms:

The inner integral is the definition of G(s) = ℬ{g(t)}. Factoring G(s) out leaves the definition of F(s) = ℬ{f(t)}, thus proving the theorem.

L {f * g} = (\int_{0}^{\infty} e^{- s τ} f (τ) d τ) (\int_{0}^{\infty} e^{- s u} g (u) d u) = F (s) G (s)

Result

L {f * g} = (\int_{0}^{\infty} e^{- s τ} f (τ) d τ) (\int_{0}^{\infty} e^{- s u} g (u) d u) = F (s) G (s)

Source: Kreyszig, E. (2011). Advanced Engineering Mathematics (10th ed.). John Wiley & Sons.

Free formulas

Rearrangements

Solve for F(s)G(s)

Make F(s)G(s) the subject

F (s) \cdot G (s)

Start from the Convolution Theorem (Laplace). The expression F(s)G(s) is already isolated, so the task is to identify it as the subject and present it in the target notation.

Difficulty: 1/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph typically displays the time-domain representation of two functions and their resulting convolution, often showing smooth, decaying, or oscillatory curves. These curves illustrate the 'memory' effect of a system, where the output at time t is a weighted integral of all past inputs. The interaction between the functions demonstrates how the convolution operation smooths out signals and transforms the system's response into a combined waveform.

Graph type: exponential

Why it behaves this way

Intuition

This theorem provides a powerful 'domain transformation' perspective, where a complex operation like convolution in the time domain is simplified into a straightforward algebraic multiplication in the frequency domain

L

The Laplace transform operator, which converts a function from the time domain (t) to the complex frequency domain (s).

It's like translating a signal's behavior over time into its frequency components, revealing its spectral 'fingerprint'.

f * g

The convolution integral of two functions, f(t) and g(t). It describes how the shape of one function modifies the shape of another, often representing the output of a linear system

Imagine 'blending' or 'smearing' one function across another, where the second function dictates the weighting or influence at each point.

F(s)

The Laplace transform of the function f(t), representing f(t) in the complex frequency domain.

This is the frequency-domain 'signature' of f(t), detailing its constituent frequencies rather than its time evolution.

G(s)

The Laplace transform of the function g(t), representing g(t) in the complex frequency domain.

Similar to F(s), this is the frequency-domain 'signature' of g(t).

Free study cues

Insight

Canonical usage

Ensures dimensional consistency between the Laplace transform of a convolution and the product of individual Laplace transforms, where the units of the Laplace variable 's' are inverse time.

Common confusion

Misinterpreting the units of the Laplace variable 's' as dimensionless, or incorrectly handling the units introduced by the integration in the definitions of the Laplace transform and convolution.

Unit systems

f(t)[U_f] - Unit of the function f in the time domain.

g(t)[U_g] - Unit of the function g in the time domain.

$t$ [T] - Unit of the independent time variable, typically seconds (s) in SI.

$s$ [T]^-1 - Unit of the Laplace variable (complex frequency), typically s^-1 or rad/s in SI.

$F (s) = L {f (t)}$ [U_f][T] - The Laplace transform integral \int_0^\infty f(t)e^{-st} dt introduces a unit of time from 'dt'.

$G (s) = L {g (t)}$ [U_g][T] - The Laplace transform integral \int_0^\infty g(t)e^{-st} dt introduces a unit of time from 'dt'.

(f*g)(t)[U_f][U_g][T] - The convolution integral \int_0^t f(\tau)g(t-\tau) d\tau introduces a unit of time from 'd\tau'.

$L {f * g}$ [U_f][U_g][T]^2 - The Laplace transform of the convolution (f*g)(t) introduces an additional unit of time, resulting in ([U_f][U_g][T]) * [T].

One free problem

Practice Problem

Given the individual transforms F(s) = 4 and G(s) = 8, calculate the Laplace transform of the convolution (f * g)(t).

F(s)4 Transform of the first function

G(s)8 Transform of the second function

Solve for: result

Hint: According to the theorem, the transform of the convolution is simply the product of the individual transforms.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In signal processing, the output of a system is the convolution of its input signal and its impulse response; this theorem allows us to find the output using multiplication in the s-domain.

Study smarter

Tips

The convolution f * g is defined as the integral from 0 to t of f(τ)g(t-τ) dτ.
Remember that convolution is commutative, meaning f * g = g * f.

Avoid these traps

Common Mistakes

Confusing convolution f*g with the pointwise product f(t)g(t).
Forgetting that the theorem only applies if the transforms F(s) and G(s) exist for the same region of convergence.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This derivation demonstrates that the Laplace transform of the convolution of two functions is equivalent to the product of their individual Laplace transforms.

Essential for solving non-homogeneous differential equations and analyzing linear time-invariant (LTI) systems.

It converts the complex operation of convolution in the time domain into simple algebraic multiplication in the frequency (s) domain.

Confusing convolution f*g with the pointwise product f(t)g(t). Forgetting that the theorem only applies if the transforms F(s) and G(s) exist for the same region of convergence.

In signal processing, the output of a system is the convolution of its input signal and its impulse response; this theorem allows us to find the output using multiplication in the s-domain.

The convolution f * g is defined as the integral from 0 to t of f(τ)g(t-τ) dτ. Remember that convolution is commutative, meaning f * g = g * f.

References

Sources

Advanced Engineering Mathematics
Wikipedia: Laplace transform
Differential Equations with Boundary-Value Problems by Dennis G. Zill
Dennis G. Zill, Warren S. Wright. Differential Equations with Boundary-Value Problems.
Erwin Kreyszig. Advanced Engineering Mathematics.
Wikipedia: Convolution theorem
Kreyszig, Advanced Engineering Mathematics
Boyce, DiPrima, and Meade, Elementary Differential Equations and Boundary Value Problems

Overview

Variables

Derivation

Start with the definition of the Laplace transform of a convolution:

Change the order of integration:

Perform a substitution in the inner integral:

Recognize the Laplace transforms:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Fourier Transform (Continuous)

Moment Generating Function (MGF)

Frequently Asked Questions

Sources